Giải thích các bước giải:
a) `x-3sqrtx=10`
`=>x-3sqrtx-10=0`
`=>(x-5sqrtx)+(2sqrtx-10)=0`
`=>sqrtx(sqrtx-5)+2(sqrtx-5)=0`
`=>(sqrtx+2)(sqrtx-5)=0`
Mà `sqrtx>=0=>sqrtx+2>0`
`=>sqrtx-5=0`
`=>sqrtx=5`
`=>x=25`
b) `sqrt(x-1)+sqrt(4x-4)-sqrt(25x-25)+2=0`
`=>sqrt(x-1)+2sqrt(x-1)-5sqrt(x-1)+2=0`
`=>-2sqrt(x-1)+2=0`
`=>sqrt(x-1)=1`
`=>x-1=1`
`=>x=2`
c) `sqrt(x^2-2x+1)=x^2-1`
`=>sqrt((x-1)^2)-x^2+1=0`
`=>|x-1|-x^2+1=0`
`=>`\(\left[ \begin{array}{l}x-1-x^2+1=0\\1-x-x^2+1=0\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x-x^2=0\\-x^2-x+2=0\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x(1-x)=0\\-(x-1)(x+2)=0\end{array} \right.\) `=>`\(\left[ \begin{array}{l}\left[ \begin{array}{l}x=0\\1-x=0\end{array} \right.\\\left[ \begin{array}{l}x-1=0\\x+2=0\end{array} \right.\end{array} \right.\) `=>`\(\left[ \begin{array}{l}\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\\\left[ \begin{array}{l}x=1\\x=-2\end{array} \right.\end{array} \right.\)
Vậy `x in{0;1;-2}.`
